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Prove that

$$\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}(k+1)^n=0\;.$$

I tried finding a combinatorial interpretation but to no avail. Here is a combinatorial statement, however crappy. Suppose we have $n$ blanks on a sheet of paper. We choose $k$ (at least one) of them. Afterwards, we fill all blanks except for the leftmost one with an integer from $1$ to $k$. Show that the number of ways to do this with $k$ even is the same as the number of ways to do this with $k$ odd.

I don't see how it is the same with the identity posted in the comments. I did not introduce any factorials.

Martin Sleziak
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1 Answers1

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This sequence shows up quite frequently here. Suppose we seek to evaluate $$\sum_{k=0}^n {n+1\choose k+1} (-1)^k (k+1)^q$$ whith $q$ a non-negative integer.

Introduce the integral representation $$(k+1)^q = \frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \exp((k+1)z) \; dz.$$

This gives for the sum $$\frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \sum_{k=0}^n {n+1\choose k+1} (-1)^k \exp((k+1)z) \; dz \\ = -\frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \sum_{k=0}^n {n+1\choose k+1} (-1)^{k+1} \exp((k+1)z) \; dz.$$

This is $$-\frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \left((1-\exp(z))^{n+1}-1\right) \; dz$$ which evaluates to $$-q! [z^q] (1-\exp(z))^{n+1} \\ = -q! \times (n+1)! \times (-1)^{n+1} [z^q] \frac{(\exp(z)-1)^{n+1}}{(n+1)!}.$$ This is $$(n+1)! \times (-1)^{n} \times {q\brace n+1}.$$

The above is zero when $q\le n$ which was to be shown.

Remark. Here we have used the combinatorial class of set partitions which is given by

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\mathcal{U}\times\textsc{SET}_{\ge 1}(\mathcal{Z})),$$

producing the generating function $$\exp(u(\exp(z)-1))$$ and in particular $${n\brace k} = n! [z^n] \frac{(\exp(z)-1)^k}{k!}.$$

A similar calculation can be found at this MSE link.

Marko Riedel
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